Question

According to Moseley, for $K_{\alpha }$ line$\sqrt{v}$ $=a(z-b)$ where $a=5\times {10}^7$ Hz and $b=1$. For a given element, the wavelength of $K_{\alpha }$ line is $\frac{1}{3}{A}^0$. Then the atomic number of the element is

• A. 51
• B. 61
• C. 41
• D. 55

Answer 1

The correct option is B 61

$\sqrt{V}=a(z-b)$
$V=a^2(z-b{)}^2$

$\frac{C}{\lambda }=a^2(z-b{)}^2$

$\frac{3\times {10}^8}{1/3\times {10}^{-10}}=25\times {10}^{14}(z-1{)}^2$

$3^2\times {10}^{18}=5^2\times {10}^{14}(z-1{)}^2$

$\frac{3}{5}\times {10}^2=(z-1)$

$z-1=\frac{3}{5}\times {10}^2$

$z=61$

So, the answer is option (B).